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Solving derivates, getting the minimum of a function, and helpful documentation of the deriv function

3 messages · Sorkin, John, Roy Mendelssohn - NOAA Federal, Rolf Turner

Sorkin, John
# 
I am trying to find the minimum of a linear function:

y <- (-0.0263*b) + (0.0010*B^2)

I am having GREAT difficulty with the documentation of the deriv function. I have (after playing for two-hours) been able to get the following to work:

zoop <- deriv(expression((-0.0263*B)+(0.0010*B^2)),"B",func=TRUE)
class(zoop)
zoop(2)

which appears to give me the value of the derivative of my expression w.r.t. B
(I am not certain what the func arugment does, but it appears to be necessary)

Following what one learns in calculus 1, I now need to set the derivative equal to 0 and solve for B. I have no idea how to do this

Can someone point me in the right direction. Additionally can someone suggest documentation for deriv that is easily intelligible to someone who wants to learn how to use the function, rather that documentation that helps one who is already familiar with the function. (I have a need for derivatives that is beyond finding the minimum of a function)

Thank you
John

P.S. Please don?t flame. I spent a good deal of time looking at documentation and searching the internet. There may be something on line, but I clearly am not using the correct search terms.
Roy Mendelssohn - NOAA Federal
# 
Hi John:

Can I ask if this is the specific problem you are after,  or a test for  more general problem?  If the former,  the derivative is

 -0.0263 + 0.002 * B

so the solution for B is:

B = (0263)/0.002

If you are after a more general way fo doing this:

?solve

-Roy
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Roy Mendelssohn
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NOAA/NMFS
Environmental Research Division
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Rolf Turner
# 
On Sat, 29 Aug 2020 21:15:56 +0000
"Sorkin, John" <jsorkin at som.umaryland.edu> wrote:

            

      
      
      
    

        
Quadratic function???

            

      
      
      
    

        
It causes deriv() to return a *function* rather than an *expression*.

            

      
      
      
    

        
Couple of things that you could play around with.

y <- expression(-0.0263*B + 0.0010*B^2)
z <- deriv(y,"B",func=TRUE)
f <- function(x,z){as.vector(attr(z(x),"gradient"))}

(1) uniroot(f,c(5,15),z=z)$root
# 13.15 --- right answer!!! :-)

(2) library(polynom) # You may need to install this package.
    p <- poly.calc(x=1:2,y=f(1:2,z=z))
    polyroot(p)
# 13.15+0i You can get rid of the extraneous imaginary part
# by using Re(polyroot(p))

HTH

cheers,

Rolf

P.S. It's irritating the way that one has to fiddle about in order to
get a function that returns the value of the derivative, rather than the
value of the function being differentiated!

R.